$\begin{bmatrix} -2 & 1 &0\\ 0 & -2 &1\\ 0 & 0 &-2 \end{bmatrix}$ , x(t)=$\begin{bmatrix}x_1(t)\\ x_2(t) \\ x_3(t)\end{bmatrix}$ and


Then any solution of the first order system of the ordinary differential equation

$\left\{ \begin{array}{rcl} x'(t)=Ax(t)\\ x(0)=x_0 \end{array}\right. \;\;$


  1. $\lim _{t\rightarrow \infty |x(t)|=0}$

  2. $\lim _{t\rightarrow \infty |x(t)|=\infty}$

  3. $\lim _{t\rightarrow \infty |x(t)|=2}$

  4. $\lim _{t\rightarrow \infty |x(t)|=12}$
  • 1
    $\begingroup$ Compute $e^{At}x_0$ and you will find the answer. Note that $A$ is a jordan-matrix. $\endgroup$ – Fakemistake May 31 '17 at 7:59

Since $-2$ is the only eigenvalue of $A$, the general solution of $x'(t)=Ax(t)$ has the form


where $c_1,c_2$ and $c_3$ are constants and $v_j(t)$ is a vector in $ \mathbb R^3$ whose coordinates are polynomials with grade $ \le j-1$.

Hence $|x(t)| \to 0$ for $t \to \infty$.


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